JAMB - Physics (1983 - No. 1)
In a resonance tube experiment, a tube fixed length is closed at one end and several tuning forks of increasing frequency used to obtain resonance at the open end. If the tuning fork with the lowest frequency which gave resonance had a frequency f1 and the next tuning fork to give resonance had a frequency f2, find the ratio \(\frac{f^2}{f^1}\).
8
3
2
\(\frac{1}{2}\)
\(\frac{1}{3}\)
Wyjaśnienie
\(\lambda_1\) = 4L\(_1\) and \(\lambda_2\) = \(\frac{4L_1}{3}\)
f\(\lambda\) = velocity But f = \(\frac{V}{\lambda}\)
F\(_1\) = \(\frac{V}{4L_1}\) and F\(_2\) = \(\frac{3V}{4L_1}\)
\(\frac{F_2}{F_1}\) = \(\frac{\frac{3V}{4L_1}}{\frac{V}{4L_1}}\) = 3
Note: The next frequency F\(_2\) that gives resonance will correspond to the third harmonic, which supports three-quarters of a wavelength